Minimum sum of two numbers formed from digits of an array

Given an array of digits (values are from 0 to 9), find the minimum possible sum of two numbers formed from digits of the array. All digits of given array must be used to form the two numbers.

Examples:

Input: [6, 8, 4, 5, 2, 3]
Output: 604
The minimum sum is formed by numbers 
358 and 246

Input: [5, 3, 0, 7, 4]
Output: 82
The minimum sum is formed by numbers 
35 and 047



Since we want to minimize the sum of two numbers to be formed, we must divide all digits in two halves and assign half-half digits to them. We also need to make sure that the leading digits are smaller.
We build a Min Heap with the elements of the given array, which takes O(n) worst time. Now we retrieve min values (2 at a time) of array, by polling from the Priority Queue and append these two min values to our numbers, till the heap becomes empty, i.e., all the elements of array get exhausted. We return the sum of two formed numbers, which is our required answer. Overall complexity is O(nlogn) as push() operation takes O(logn) and it’s repeated n times.

C/C++

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// C++ program to find minimum sum of two numbers
// formed from all digits in a given array.
#include<bits/stdc++.h>
using namespace std;
  
// Returns sum of two numbers formed
// from all digits in a[]
int minSum(int arr[], int n)
{
    // min Heap
    priority_queue <int, vector<int>, greater<int> > pq;
  
    // to store the 2 numbers formed by array elements to
    // minimize the required sum
    string num1, num2;
  
    // Adding elements in Priority Queue
    for(int i=0; i<n; i++)
        pq.push(arr[i]);
  
    // checking if the priority queue is non empty
    while(!pq.empty())
    {
        // appending top of the queue to the string
        num1+=(48 + pq.top());
        pq.pop();
        if(!pq.empty())
        {
            num2+=(48 + pq.top());
            pq.pop();
        }
    }
  
    // converting string to integer
    int a = atoi(num1.c_str());
    int b = atoi(num2.c_str());
  
    // returning the sum
    return a+b;
}
  
int main()
{
    int arr[] = {6, 8, 4, 5, 2, 3};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout<<minSum(arr, n)<<endl;
    return 0;
}
// Contributed By: Harshit Sidhwa

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Java














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// Java program to find minimum sum of two numbers 


// formed from all digits in a given array. 


import java.util.PriorityQueue; 


  


class MinSum 


{ 


    // Returns sum of two numbers formed 


    // from all digits in a[] 


    public static long solve(int[] a) 


    { 


        // min Heap 


        PriorityQueue<Integer> pq = new PriorityQueue<Integer>(); 


  


        // to store the 2 numbers formed by array elements to 


        // minimize the required sum 


        StringBuilder num1 = new StringBuilder(); 


        StringBuilder num2 = new StringBuilder(); 


  


        // Adding elements in Priority Queue 


        for (int x : a) 


            pq.add(x); 


  


        // checking if the priority queue is non empty 


        while (!pq.isEmpty()) 


        { 


            num1.append(pq.poll()+ ""); 


            if (!pq.isEmpty()) 


                num2.append(pq.poll()+ ""); 


        } 


  


        // the required sum calculated 


        long sum = Long.parseLong(num1.toString()) + 


                   Long.parseLong(num2.toString()); 


  


        return sum; 


    } 


  


    // Driver code 


    public static void main (String[] args) 


    { 


        int arr[] = {6, 8, 4, 5, 2, 3}; 


        System.out.println("The required sum is "+ solve(arr)); 


    } 


} 



















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Output:


The required sum is 604



Anothor method:

We can follow another approach also like this, as we need two numbers such that their sum is minimum, then we would also need two minimum numbers. If we arrange our array in ascending order then we can two digits that will form the smallest numbers,

e.g, 2 3 4 5 6 8, now we can get two numbers starting from 2 and 3. Now first part is done. Now we have to form such that they would contain small digits, i.e pick digits alternatively from array extend your two numbers.

i.e 246, 358. Now if we see analyze this, then we can pick even indexed numbers for num1 and odd number for num2.


Below is the implementation:




C++














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// C++ program to find minimum sum of two numbers 


// formed from all digits in a given array. 


#include<bits/stdc++.h> 


using namespace std; 


  


// Returns sum of two numbers formed 


// from all digits in a[] 


int minSum(int a[], int n){ 


      


    // sort the elements 


    sort(a,a+n); 


      


    int num1 = 0; 


    int num2 = 0; 


    for(int i = 0;i<n;i++){ 


        if(i%2==0) 


            num1 = num1*10+a[i]; 


        else num2 = num2*10+a[i]; 


    } 


    return num2+num1; 


} 


  


// Driver code 


int main() 


{ 


    int arr[] = {5, 3, 0, 7, 4};  


    int n = sizeof(arr)/sizeof(arr[0]);  


    cout<<"The required sum is  "<<minSum(arr, n)<<endl;  


    return 0; 


} 



















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Java














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import java.util.Arrays; 


//Java program to find minimum sum of two numbers 


//formed from all digits in a given array. 


public class AQRQ { 


  


    //Returns sum of two numbers formed 


    //from all digits in a[] 


    static int minSum(int a[], int n){ 


        


     // sort the elements 


     Arrays.sort(a); 


        


     int num1 = 0; 


     int num2 = 0; 


     for(int i = 0;i<n;i++){ 


         if(i%2==0) 


             num1 = num1*10+a[i]; 


         else num2 = num2*10+a[i]; 


     } 


     return num2+num1; 


    } 


  


    //Driver code 


    public static void main(String[] args) { 


          


  


         int arr[] = {5, 3, 0, 7, 4};  


         int n = arr.length; 


         System.out.println("The required sum is  " + minSum(arr, n)); 


  


    } 


  


} 



















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Python3














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# Python 3 program to find minimum  


# sum of two numbers formed  


# from all digits in an given array 


  


# Returns sum of two numbers formed  


# from all digits in a[] 


def minSum(a, n): 


      


    # sorted the elements 


    a = sorted(a) 


    num1, num2 = 0, 0


      


    for i in range(n): 


        if i % 2 == 0: 


            num1 = num1 * 10 + a[i] 


        else: 


            num2 = num2 * 10 + a[i] 


      


    return num2 + num1      


      


# Driver code 


arr = [5, 3, 0, 7, 4] 


n = len(arr) 


print("The required sum is", 


             minSum(arr, n)) 


      


# This code is contributed 


# by Mohit kumar 29 



















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C#














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// C# a program to find minimum sum of two numbers 


//formed from all digits in a given array. 


  


using System; 


  


public class GFG{ 


      


    //Returns sum of two numbers formed 


    //from all digits in a[] 


    static int minSum(int []a, int n){ 


      


    // sort the elements 


    Array.Sort(a); 


      


    int num1 = 0; 


    int num2 = 0; 


    for(int i = 0;i<n;i++){ 


        if(i%2==0) 


            num1 = num1*10+a[i]; 


        else num2 = num2*10+a[i]; 


    } 


    return num2+num1; 


    } 


  


    //Driver code 


    static public void Main (){ 


        int []arr = {5, 3, 0, 7, 4};  


        int n = arr.Length; 


        Console.WriteLine("The required sum is " + minSum(arr, n)); 


    } 


//This code is contributed by ajit.     


} 



















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PHP














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<?php 


// PHP program to find minimum sum  


// of two numbers formed from all  


// digits in a given array.  


  


// Returns sum of two numbers formed  


// from all digits in a[]  


function minSum($a, $n) 


{  


      


    // sort the elements  


    sort($a);  


      


    $num1 = 0;  


    $num2 = 0;  


    for($i = 0; $i < $n; $i++) 


    {  


        if($i % 2 == 0)  


            $num1 = $num1 * 10 + $a[$i];  


        else $num2 = $num2 * 10 + $a[$i];  


    }  


    return ($num2 + $num1);  


}  


  


// Driver code  


$arr = array(5, 3, 0, 7, 4);  


$n = sizeof($arr);  


echo "The required sum is ",  


     minSum($arr, $n), "\n";  


  


// This Code is Contributed by ajit 


?> 



















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Output:


The required sum is 82





Time Complexity : O(nLogN)


This article is contributed by Prakhar. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.


Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



          
          
          
          

            

    

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